Solving inequations so that $x \rightarrow a^x$ 
The exponential function $x \rightarrow a^x$, with $0 < a < 1$ is
  decrescent in $\mathbb{R}$.
Taking this property in consideration, solve the following
  inequalities:
a. $(\frac{1}{3})^{2x} \le (\frac{1}{3})^{x+1}$
b. $0.1^{2-x}>0.1^{3x}$

I did:
a.$$2x \le n+1 \Leftrightarrow 2x-x \le 1 \Leftrightarrow x \le 1$$
b.$$2-x > 3x \Leftrightarrow \frac{2-x}{3} > x \Leftrightarrow -\frac{x}{3}-x > \frac{2}{3} \Leftrightarrow \frac{x}{3}+x < \frac{2}{3} \Leftrightarrow \frac{4x}{3} < \frac{2}{3} \Leftrightarrow 4x < 2 \Leftrightarrow x < \frac{1}{2}$$
But my book says the solution is $x \ge 1$ for a. and $x>\frac{1}{2}$ for b.
What did I do wrong? Did I overlook something regarding the given property?
 A: 
When the sides of an inequality are multiplied (or divided) by a negative number, the inequality turns to the opposite side.

It appears you took the $\log$ of the sides to write your inequalities. in the first case, for instance,
$$\log\left(\left(\frac{1}{3}\right)^{2x}\right) \le \log\left(\left(\frac{1}{3}\right)^{x+1}\right)$$
$$2x\log\left(\frac{1}{3}\right) \le (x+1)\log\left(\frac{1}{3}\right)$$
Dividing the sides by $\log\left(\frac{1}{3}\right)$ $\color{red}{\text{which is negative}}$ yields:
$$2x\ge x+1\Rightarrow x\ge1$$
Since $\frac{1}{3}<1$ and $0.1<1$, this will flip the inequalities (since the result of $\log$ is negative). So the result is $x\ge1$ and $x>\frac{1}{2}$.
A: It seems like you have done it! Just correct your inequalities (directions). Just in case, here are the solutions:
Let's just use what we are given: $f(x)=a^x$ is decreasing if $0<a<1$.
$\textbf{a}$: Consider the function $f(x)=(\frac{1}{3})^x$. The given information says that if we plug in something greater, then the image will decrease. We have $f(2x)\leq f(x+1) \iff 2x \geq x+1$. Thus $x\geq 1$. To be more rigorious, actually what we have used here is as follows: A function $f$ is decreasing on an interval $I$ $\iff$ $\forall x_{1},x_{2} \in I$, if  $x_{1}\geq x_{2}$, then $f(x_{1}) \leq f(x_{2})$. 
$\textbf{b}$: Precisely the same case, since $0<0.1<1$. Therefore, $3x>2-x \implies x>\frac{1}{2}$.
I hope this is useful,
TT.
